Integrand size = 21, antiderivative size = 123 \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\frac {(12-n) \cos ^4(c+d x) (a+a \sec (c+d x))^{3+n}}{20 a^3 d}-\frac {\cos ^5(c+d x) (a+a \sec (c+d x))^{3+n}}{5 a^3 d}+\frac {\left (32-13 n+n^2\right ) \operatorname {Hypergeometric2F1}(4,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{20 a^3 d (3+n)} \]
1/20*(12-n)*cos(d*x+c)^4*(a+a*sec(d*x+c))^(3+n)/a^3/d-1/5*cos(d*x+c)^5*(a+ a*sec(d*x+c))^(3+n)/a^3/d+1/20*(n^2-13*n+32)*hypergeom([4, 3+n],[4+n],1+se c(d*x+c))*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)
Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=-\frac {\left ((3+n) \cos ^4(c+d x) (-12+n+4 \cos (c+d x))-\left (32-13 n+n^2\right ) \operatorname {Hypergeometric2F1}(4,3+n,4+n,1+\sec (c+d x))\right ) (1+\sec (c+d x))^3 (a (1+\sec (c+d x)))^n}{20 d (3+n)} \]
-1/20*(((3 + n)*Cos[c + d*x]^4*(-12 + n + 4*Cos[c + d*x]) - (32 - 13*n + n ^2)*Hypergeometric2F1[4, 3 + n, 4 + n, 1 + Sec[c + d*x]])*(1 + Sec[c + d*x ])^3*(a*(1 + Sec[c + d*x]))^n)/(d*(3 + n))
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4361, 27, 100, 27, 87, 75}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sin ^5(c+d x) (a \sec (c+d x)+a)^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^5 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4361 |
| \(\displaystyle -\frac {\int a^2 \cos ^6(c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{n+2}d(-\sec (c+d x))}{a^4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \cos ^6(c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{n+2}d(-\sec (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle -\frac {\frac {\int -a \cos ^5(c+d x) (-n-5 \sec (c+d x)+12) (\sec (c+d x) a+a)^{n+2}d(-\sec (c+d x))}{5 a}+\frac {\cos ^5(c+d x) (a \sec (c+d x)+a)^{n+3}}{5 a}}{a^2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {1}{5} \int -\cos ^5(c+d x) (-n-5 \sec (c+d x)+12) (\sec (c+d x) a+a)^{n+2}d(-\sec (c+d x))+\frac {\cos ^5(c+d x) (a \sec (c+d x)+a)^{n+3}}{5 a}}{a^2 d}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {1}{4} \left (n^2-13 n+32\right ) \int \cos ^4(c+d x) (\sec (c+d x) a+a)^{n+2}d(-\sec (c+d x))-\frac {(12-n) \cos ^4(c+d x) (a \sec (c+d x)+a)^{n+3}}{4 a}\right )+\frac {\cos ^5(c+d x) (a \sec (c+d x)+a)^{n+3}}{5 a}}{a^2 d}\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle -\frac {\frac {1}{5} \left (-\frac {\left (n^2-13 n+32\right ) (a \sec (c+d x)+a)^{n+3} \operatorname {Hypergeometric2F1}(4,n+3,n+4,\sec (c+d x)+1)}{4 a (n+3)}-\frac {(12-n) \cos ^4(c+d x) (a \sec (c+d x)+a)^{n+3}}{4 a}\right )+\frac {\cos ^5(c+d x) (a \sec (c+d x)+a)^{n+3}}{5 a}}{a^2 d}\) |
-(((Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(3 + n))/(5*a) + (-1/4*((12 - n)*C os[c + d*x]^4*(a + a*Sec[c + d*x])^(3 + n))/a - ((32 - 13*n + n^2)*Hyperge ometric2F1[4, 3 + n, 4 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3 + n) )/(4*a*(3 + n)))/5)/(a^2*d))
3.2.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _), x_Symbol] :> Simp[-(f*b^(p - 1))^(-1) Subst[Int[(-a + b*x)^((p - 1)/2 )*((a + b*x)^(m + (p - 1)/2)/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[ {a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{5}d x\]
\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]
\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int {\sin \left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]